y''-4y=2; y1=e^(-2x)
i dont see how my text got yp=1/2
ok so this is what i have so far....
let y2=ue^(-2x)
so
so
i have the yc part i just dont know how to get the yp..what am i missing or not getting??
x''(t)-4x(t)=2. First find the homogeneous x''(t)-4x(t)=0, characteristic equation m^2-4=0, m1/2=+-2, xh=C1*e^(m1*t)+C2*e^(m2*t)=C1e^(2*t)+C2*e^(-2*t). As we see that a non homogeneous part in start equation is 2, xp must be constant. x=xh+xp. Substitute this in start equation (C1*e^2t+C2*e(-2t)+xp)''-4(C1*e^2t+C2*e(-2t)+xp)=2, xp''=0, -4xp=2, xp=-1/2.
Kind regards,
derdack, yes, the problem can be solved that way, but I don't believe it is a god idea to advise a person to ignore the teacher's or text's instructions. Slapmaxwell1 says he was instructed to use "reduction of order". It is not at all uncommon to ask a person to solve a simple problem using a more complicated technique in order to practice that technique.
Slapmaxwell1, yes, letting $\displaystyle y= ue^{2x}$ reduces the left side of the equation to $\displaystyle e^{-2x}(u''- 4u')$ but you appear to have left out the right side of the equation. You have $\displaystyle e^{-2x}(u''- 4u')= 2$ so that $\displaystyle u''- 4u'= 2e^{-2x}$ and, letting w= u', $\displaystyle w'- 4w= 2e^{-2x}$.
That is now a first order equation linear equation. It has $\displaystyle e^{-4x}$ as integrating factor:
$\displaystyle e^{-4x}w'- 4e^{-4x}w= (e^{-4x}w)'= 2e^{-4x}$
ok ok i see it now.
if i let $\displaystyle y_{p}=A$ then $\displaystyle y'=0 , y''=0$
ok so $\displaystyle 0-4A=2$ so $\displaystyle A=\frac{-1}{2}$
there for $\displaystyle y_{p}=\frac{-1}{2}$ so $\displaystyle y=y_{c}+y_{p}$ because i found Yc I would just put it all together for my final answer. thanks again!